How To Find Terminal Voltage

21 Circuits and DC Instruments

162 21.two Electromotive Forcefulness: Concluding Voltage

Summary

• Compare and contrast the voltage and the electromagnetic force of an electric ability source.
• Describe what happens to the terminal voltage, current, and power delivered to a load every bit internal resistance of the voltage source increases (due to aging of batteries, for example).
• Explain why it is beneficial to use more than ane voltage source continued in parallel.

When you lot forget to turn off your car lights, they slowly dim as the battery runs down. Why don't they just glimmer off when the bombardment's energy is gone? Their gradual dimming implies that battery output voltage decreases as the battery is depleted.

Furthermore, if yous connect an excessive number of 12-V lights in parallel to a automobile battery, they volition be dim even when the battery is fresh and even if the wires to the lights have very low resistance. This implies that the battery's output voltage is reduced by the overload.

The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts—a source of electrical energy and an internal resistance. Allow us examine both.

Electromotive Force

You can think of many unlike types of voltage sources. Batteries themselves come in many varieties. There are many types of mechanical/electrical generators, driven by many different energy sources, ranging from nuclear to wind. Solar cells create voltages directly from light, while thermoelectric devices create voltage from temperature differences.

A few voltage sources are shown in Figure 1. All such devices create a potential difference and can supply electric current if connected to a resistance. On the small scale, the potential difference creates an electrical field that exerts force on charges, causing current. We thus use the proper name electromotive force, abbreviated emf.

Emf is not a forcefulness at all; information technology is a special type of potential divergence. To be precise, the electromotive force (emf) is the potential difference of a source when no current is flowing. Units of emf are volts.

Figure 1. A variety of voltage sources (clockwise from top left): the Brazos Wind Farm in Fluvanna, Texas (credit: Leaflet, Wikimedia Eatables); the Krasnoyarsk Dam in Russia (credit: Alex Polezhaev); a solar farm (credit: U.S. Department of Energy); and a group of nickel metal hydride batteries (credit: Tiaa Monto). The voltage output of each depends on its construction and load, and equals emf but if in that location is no load.

Electromotive force is directly related to the source of potential divergence, such every bit the detail combination of chemicals in a battery. However, emf differs from the voltage output of the device when current flows. The voltage across the terminals of a bombardment, for instance, is less than the emf when the battery supplies current, and it declines further every bit the battery is depleted or loaded downwardly. All the same, if the device's output voltage tin exist measured without drawing current, and so output voltage will equal emf (even for a very depleted battery).

Internal Resistance

As noted earlier, a 12-V truck battery is physically larger, contains more accuse and energy, and tin can deliver a larger electric current than a 12-V motorcycle bombardment. Both are lead-acrid batteries with identical emf, but, considering of its size, the truck battery has a smaller internal resistance [latex]\boldsymbol{r}[/latex]. Internal resistance is the inherent resistance to the flow of current within the source itself.

Figure 2 is a schematic representation of the two key parts of whatever voltage source. The emf (represented by a script Due east in the effigy) and internal resistance [latex]\boldsymbol{r}[/latex] are in serial. The smaller the internal resistance for a given emf, the more current and the more than ability the source tin supply.

Effigy 2. Whatsoever voltage source (in this instance, a carbon-zinc dry cell) has an emf related to its source of potential deviation, and an internal resistance r related to its construction. (Notation that the script E stands for emf.). Also shown are the output terminals across which the last voltage V is measured. Since V = emf − Ir , terminal voltage equals emf merely if there is no current flowing.

The internal resistance [latex]\boldsymbol{r}[/latex] can bear in complex ways. As noted, [latex]\boldsymbol{r}[/latex] increases as a battery is depleted. But internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and fifty-fifty its history. The internal resistance of rechargeable nickel-cadmium cells, for case, depends on how many times and how deeply they have been depleted.

Things Great and Small: The Submicroscopic Origin of Battery Potential

Various types of batteries are bachelor, with emfs determined by the combination of chemicals involved. We can view this equally a molecular reaction (what much of chemical science is about) that separates charge.

The atomic number 82-acid bombardment used in cars and other vehicles is one of the nearly mutual types. A unmarried cell (1 of six) of this battery is seen in Figure iii. The cathode (positive) terminal of the jail cell is connected to a atomic number 82 oxide plate, while the anode (negative) final is connected to a atomic number 82 plate. Both plates are immersed in sulfuric acid, the electrolyte for the organisation.

Figure 3. Artist's formulation of a lead-acid cell. Chemical reactions in a lead-acid cell separate charge, sending negative charge to the anode, which is continued to the atomic number 82 plates. The lead oxide plates are continued to the positive or cathode terminal of the cell. Sulfuric acrid conducts the charge too as participating in the chemical reaction.

The details of the chemical reaction are left to the reader to pursue in a chemistry text, only their results at the molecular level aid explain the potential created by the battery. Figure 4 shows the issue of a single chemical reaction. Two electrons are placed on the anode, making information technology negative, provided that the cathode supplied two electrons. This leaves the cathode positively charged, because it has lost two electrons. In brusque, a separation of charge has been driven by a chemical reaction.

Note that the reaction will non accept identify unless there is a complete circuit to let two electrons to exist supplied to the cathode. Under many circumstances, these electrons come from the anode, flow through a resistance, and return to the cathode. Note as well that since the chemic reactions involve substances with resistance, it is not possible to create the emf without an internal resistance.

Effigy 4. Artist'south conception of two electrons existence forced onto the anode of a jail cell and 2 electrons being removed from the cathode of the prison cell. The chemical reaction in a lead-acrid battery places two electrons on the anode and removes two from the cathode. It requires a airtight circuit to proceed, since the 2 electrons must be supplied to the cathode.

Why are the chemicals able to produce a unique potential difference? Quantum mechanical descriptions of molecules, which take into business relationship the types of atoms and numbers of electrons in them, are able to predict the energy states they tin have and the energies of reactions between them.

In the example of a lead-acid battery, an energy of two eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy divided past charge: [latex]\boldsymbol{V = \frac{P_\textbf{E}}{q}}[/latex]. An electron volt is the energy given to a unmarried electron by a voltage of ane Five. So the voltage hither is 2 V, since ii eV is given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a dissimilar free energy and, hence, a different voltage.

Terminal Voltage

The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage [latex]\boldsymbol{V}[/latex]. Terminal voltage is given by

[latex]\boldsymbol{Five = \textbf{emf} - Ir} ,[/latex]

where [latex]\boldsymbol{r}[/latex] is the internal resistance and [latex]\boldsymbol{I}[/latex] is the electric current flowing at the time of the measurement.

[latex]\boldsymbol{I}[/latex] is positive if electric current flows away from the positive terminal, as shown in Effigy 2. You tin see that the larger the electric current, the smaller the terminal voltage. And information technology is likewise true that the larger the internal resistance, the smaller the terminal voltage.

Suppose a load resistance [latex]\boldsymbol{R_{\textbf{load}}}[/latex] is continued to a voltage source, as in Figure v. Since the resistances are in series, the total resistance in the excursion is [latex]\boldsymbol{R_{\textbf{load}} + r}[/latex]. Thus the current is given by Ohm's police to exist

[latex]\boldsymbol{I =}[/latex] [latex]\boldsymbol{\frac{\textbf{emf}}{R_{\textbf{load}} + r}}[/latex].

Figure v. Schematic of a voltage source and its load R _load . Since the internal resistance r is in series with the load, it can significantly affect the concluding voltage and current delivered to the load. (Note that the script E stands for emf.)

We meet from this expression that the smaller the internal resistance [latex]\boldsymbol{r}[/latex], the greater the electric current the voltage source supplies to its load [latex]\boldsymbol{R_{\textbf{load}}}[/latex]. As batteries are depleted, [latex]\boldsymbol{r}[/latex] increases. If [latex]\boldsymbol{r}[/latex] becomes a significant fraction of the load resistance, then the current is significantly reduced, every bit the following case illustrates.

Example 1: Calculating Terminal Voltage, Ability Dissipation, Current, and Resistance: Terminal Voltage and Load

A certain battery has a 12.0-Five emf and an internal resistance of [latex]\boldsymbol{0.100 \;\Omega}[/latex]. (a) Calculate its last voltage when connected to a [latex]\boldsymbol{10.0 - \;\Omega}[/latex] load. (b) What is the terminal voltage when connected to a [latex]\boldsymbol{0.500 - \;\Omega}[/latex] load? (c) What power does the [latex]\boldsymbol{0.500 - \;\Omega}[/latex] load dissipate? (d) If the internal resistance grows to [latex]\boldsymbol{0.500 \;\Omega}[/latex], find the current, final voltage, and power dissipated by a [latex]\boldsymbol{0.500 - \;\Omega}[/latex] load.

Strategy

The analysis in a higher place gave an expression for current when internal resistance is taken into business relationship. In one case the current is found, the terminal voltage can be calculated using the equation [latex]\boldsymbol{5 = \textbf{emf} - Ir}[/latex]. One time electric current is found, the power dissipated by a resistor tin also be found.

Solution for (a)

Entering the given values for the emf, load resistance, and internal resistance into the expression above yields

[latex]\boldsymbol{I =}[/latex] [latex]\boldsymbol{\frac{ \textbf{emf}}{R_{\textbf{load} + r}}}[/latex] [latex]\boldsymbol{=}[/latex] [latex]\boldsymbol{\frac{12.0 \;\textbf{Five}}{ten.1 \;\Omega}}[/latex] [latex]\boldsymbol{= i.188 \;\textbf{A}}.[/latex]

Enter the known values into the equation [latex]\boldsymbol{V = \textbf{emf} - Ir}[/latex] to go the terminal voltage:

[latex]\brainstorm{assortment}{r @{{}={}} l} \boldsymbol{5} & \boldsymbol{\textbf{emf} - Ir = 12.0 \;\textbf{Five} - (1.188 \;\textbf{A})(0.100 \;\Omega)} \\[1em] & \boldsymbol{eleven.9 V}. \end{array}[/latex]

Discussion for (a)

The last voltage here is only slightly lower than the emf, implying that [latex]\boldsymbol{10.0 \;\Omega}[/latex] is a light load for this detail battery.

Solution for (b)

Similarly, with [latex]\boldsymbol{R_{\textbf{load}} = 0.500 \;\Omega}[/latex], the electric current is

[latex]\boldsymbol{I =}[/latex] [latex]\boldsymbol{\frac{\textbf{emf}}{R_{\textbf{load}} + r}}[/latex] [latex]\boldsymbol{=}[/latex] [latex]\boldsymbol{\frac{12.0 \;\textbf{V}}{0.600 \;\Omega}}[/latex] [latex]\boldsymbol{= twenty.0 \;\textbf{A}} .[/latex]

The terminal voltage is now

[latex]\begin{array}{r @{{}={}} fifty} \boldsymbol{V} & \boldsymbol{\textbf{emf} - Ir = 12.0 \;\textbf{Five} - (20.0 \;\textbf{A})(0.100 \;\Omega)} \\[1em] & \boldsymbol{10.0 \;\textbf{5}}. \end{array}[/latex]

Discussion for (b)

This terminal voltage exhibits a more significant reduction compared with emf, implying [latex]\boldsymbol{0.500 \;\Omega}[/latex] is a heavy load for this battery.

Solution for (c)

The power dissipated by the [latex]\boldsymbol{0.500 - \;\Omega}[/latex] load can be found using the formula $\boldsymbol{P = I^2R} $. Entering the known values gives

[latex]\boldsymbol{P_{\textbf{load}} = I^2R_{\textbf{load}} = (20.0 \;\textbf{A})^2(0.500 \;\Omega) = 2.00 \times 10^2 \;\textbf{W}}.[/latex]

Discussion for (c)

Note that this power tin besides be obtained using the expressions [latex]\boldsymbol{\frac{5^2}{R}}[/latex] or [latex]\boldsymbol{4}[/latex], where [latex]\boldsymbol{V}[/latex] is the terminal voltage (ten.0 5 in this case).

Solution for (d)

Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as bang-up equally the load resistance. Every bit before, we beginning discover the current by entering the known values into the expression, yielding

[latex]\boldsymbol{I =}[/latex] [latex]\boldsymbol{\frac{\textbf{emf}}{R_{\textbf{load}} + r}} [latex]\boldsymbol{=}[/latex] [latex]\boldsymbol{\frac{12.0 \;\textbf{V}}{i.00 \;\Omega}}[/latex] [latex]\boldsymbol{= 12.0 \;\textbf{A}}.[/latex]

Now the last voltage is

[latex]\brainstorm{array}{r @{{}={}} l} \boldsymbol{V} & \boldsymbol{\textbf{emf} - Ir = 12.0 \;\textbf{V} - (12.0 \;\textbf{A})(0.500 \;\Omega)} \\[1em] & \boldsymbol{vi.00 \;\textbf{V}}, \finish{array}[/latex]

and the power prodigal by the load is

[latex]\boldsymbol{P_{\textbf{load}} = I^two R_{\textbf{load}} = (12.0 \;\textbf{A})^two(0.500 \;\Omega) = 72.0 \;\textbf{W}}.[/latex]

Discussion for (d)

Nosotros come across that the increased internal resistance has significantly decreased final voltage, current, and power delivered to a load.

Bombardment testers, such every bit those in Figure 6, utilize pocket-size load resistors to intentionally draw current to decide whether the concluding voltage drops beneath an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low last voltage.

Figure six. These two bombardment testers measure out concluding voltage under a load to determine the condition of a battery. The big device is being used by a U.S. Navy electronics technician to test large batteries aboard the aircraft carrier USS Nimitz and has a modest resistance that can dissipate large amounts of power. (credit: U.South. Navy photo by Lensman's Mate Airman Jason A. Johnston) The small-scale device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage. (credit: Keith Williamson)

Some batteries tin be recharged by passing a current through them in the management opposite to the current they supply to a resistance. This is washed routinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in Figure 7. The voltage output of the battery charger must be greater than the emf of the battery to reverse electric current through information technology. This will cause the terminal voltage of the battery to be greater than the emf, since [latex]\boldsymbol{V = \textbf{emf} - Ir}[/latex], and [latex]\boldsymbol{I}[/latex] is at present negative.

Figure 7. A motorcar battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential.

Multiple Voltage Sources

There are 2 voltage sources when a battery charger is used. Voltage sources continued in serial are relatively simple. When voltage sources are in serial, their internal resistances add and their emfs add algebraically. (See Effigy 8.) Series connections of voltage sources are common—for case, in flashlights, toys, and other appliances. Normally, the cells are in serial in guild to produce a larger total emf.

Merely if the cells oppose one another, such as when one is put into an appliance backward, the total emf is less, since it is the algebraic sum of the individual emfs.

A battery is a multiple connexion of voltaic cells, as shown in Effigy 9. The disadvantage of series connections of cells is that their internal resistances add. Ane of the authors in one case owned a 1957 MGA that had 2 half-dozen-V batteries in series, rather than a unmarried 12-V battery. This organization produced a large internal resistance that caused him many issues in starting the engine.

Figure 8. A series connection of two voltage sources. The emfs (each labeled with a script Eastward) and internal resistances add together, giving a total emf of emf_one + emf₂ and a total internal resistance of r _i + r ₂ .

Effigy 9. Batteries are multiple connections of individual cells, as shown in this mod rendition of an one-time print. Unmarried cells, such as AA or C cells, are normally called batteries, although this is technically incorrect.

If the series connection of ii voltage sources is made into a complete circuit with the emfs in opposition, then a current of magnitude [latex]\boldsymbol{I = \frac{\textbf{emf}_1 - \textbf{emf}_2}{r_1+r_2}}[/latex] flows. See Effigy 10, for example, which shows a circuit exactly coordinating to the bombardment charger discussed above. If 2 voltage sources in serial with emfs in the same sense are connected to a load [latex]\boldsymbol{R_{\textbf{load}}}[/latex], as in Figure 11, then [latex]\boldsymbol{I = \frac{\textbf{emf}_1 + \textbf{emf}_2}{r_1 + r_2 +R_{\textbf{load}}}}[/latex] flows.

Figure 10. These two voltage sources are connected in serial with their emfs in opposition. Current flows in the direction of the greater emf and is limited to I = (emf₁− emf₂)/(r ₁+r ₂) by the sum of the internal resistances. (Note that each emf is represented by script E in the effigy.) A battery charger continued to a battery is an instance of such a connection. The charger must take a larger emf than the bombardment to reverse current through it.

Effigy 11. This schematic represents a flashlight with two cells (voltage sources) and a single bulb (load resistance) in series. The current that flows is I = (emf₁+ emf₂)/ (r _i+ r ₂ + R _load). (Note that each emf is represented past script E in the effigy.)

Take-Domicile Experiment: Flashlight Batteries

Discover a flashlight that uses several batteries and detect new and erstwhile batteries. Based on the discussions in this module, predict the brightness of the flashlight when dissimilar combinations of batteries are used. Do your predictions match what you observe? At present identify new batteries in the flashlight and leave the flashlight switched on for several hours. Is the flashlight yet quite brilliant? Do the same with the old batteries. Is the flashlight equally bright when left on for the same length of time with sometime and new batteries? What does this say for the instance when you are limited in the number of available new batteries?

Figure 12 shows ii voltage sources with identical emfs in parallel and connected to a load resistance. In this uncomplicated case, the full emf is the same as the individual emfs. But the full internal resistance is reduced, since the internal resistances are in parallel. The parallel connection thus can produce a larger current.

Hither, [latex]\boldsymbol{I = \frac{\textbf{emf}}{(r_{\textbf{tot}} + R_{\textbf{load}})}}[/latex] flows through the load, and [latex]\boldsymbol{r_{\textbf{tot}}}[/latex] is less than those of the individual batteries. For example, some diesel-powered cars apply two 12-V batteries in parallel; they produce a total emf of 12 V but tin deliver the larger current needed to start a diesel engine.

Effigy 12. 2 voltage sources with identical emfs (each labeled by script E) continued in parallel produce the aforementioned emf simply have a smaller total internal resistance than the individual sources. Parallel combinations are ofttimes used to deliver more current. Here I = (emf)/(r _tot+R _load) flows through the load.

Animals as Electric Detectors

A number of animals both produce and detect electrical signals. Fish, sharks, platypuses, and echidnas (spiny anteaters) all notice electric fields generated by nerve activeness in casualty. Electric eels produce their own emf through biological cells (electrical organs) chosen electroplaques, which are arranged in both series and parallel as a set of batteries.

Electroplaques are flat, disk-like cells; those of the electric eel take a voltage of 0.15 5 beyond each i. These cells are unremarkably located toward the head or tail of the creature, although in the case of the electrical eel, they are institute along the entire body. The electroplaques in the South American eel are arranged in 140 rows, with each row stretching horizontally forth the body and containing five,000 electroplaques. This can yield an emf of approximately 600 Five, and a electric current of one A—deadly.

The mechanism for detection of external electric fields is like to that for producing nerve signals in the cell through depolarization and repolarization—the move of ions beyond the cell membrane. Inside the fish, weak electric fields in the water produce a current in a gel-filled canal that runs from the peel to sensing cells, producing a nerve signal. The Australian platypus, ane of the very few mammals that lay eggs, can find fields of [latex]\boldsymbol{30 \;\frac{\textbf{mV}}{\textbf{thou}}}[/latex], while sharks take been institute to exist able to sense a field in their snouts as small every bit [latex]\boldsymbol{100 \;\frac{\textbf{mV}}{\textbf{chiliad}}}[/latex] (Figure 13). Electric eels use their ain electric fields produced by the electroplaques to stun their prey or enemies.

Effigy 13. Sand tiger sharks (Carcharias taurus), like this one at the Minnesota Zoo, employ electroreceptors in their snouts to locate prey. (credit: Jim Winstead, Flickr)

Solar Prison cell Arrays

Another example dealing with multiple voltage sources is that of combinations of solar cells—wired in both serial and parallel combinations to yield a desired voltage and current. Photovoltaic generation (PV), the conversion of sunlight directly into electricity, is based upon the photoelectric effect, in which photons hitting the surface of a solar prison cell create an electric current in the prison cell.

Most solar cells are made from pure silicon—either as single-crystal silicon, or equally a thin film of silicon deposited upon a glass or metal backing. Most single cells have a voltage output of about 0.5 V, while the current output is a function of the amount of sunlight upon the cell (the incident solar radiation—the insolation). Under vivid noon sunlight, a current of about [latex]\boldsymbol{100 \;\textbf{mA/cm}^ii}[/latex] of cell surface expanse is produced past typical unmarried-crystal cells.

Individual solar cells are connected electrically in modules to see electric-energy needs. They can be wired together in series or in parallel—connected like the batteries discussed earlier. A solar-jail cell array or module commonly consists of between 36 and 72 cells, with a power output of fifty W to 140 Westward.

The output of the solar cells is directly current. For almost uses in a home, Air-conditioning is required, so a device called an inverter must exist used to catechumen the DC to AC. Whatsoever extra output can then exist passed on to the outside electrical filigree for auction to the utility.

Take-Dwelling Experiment: Virtual Solar Cells

One can gather a "virtual" solar prison cell array past using playing cards, or business organisation or index cards, to represent a solar cell. Combinations of these cards in series and/or parallel can model the required array output. Assume each card has an output of 0.5 V and a current (nether bright light) of 2 A. Using your cards, how would y'all suit them to produce an output of vi A at iii V (18 Due west)?

Suppose yous were told that y'all needed only 18 W (merely no required voltage). Would yous need more than cards to make this system?

Section Summary

• All voltage sources take two fundamental parts—a source of electrical free energy that has a feature electromotive force (emf), and an internal resistance [latex]\boldsymbol{r}[/latex].
• The emf is the potential difference of a source when no current is flowing.
• The numerical value of the emf depends on the source of potential departure.
• The internal resistance [latex]\boldsymbol{r}[/latex] of a voltage source affects the output voltage when a current flows.
• The voltage output of a device is called its terminal voltage [latex]\boldsymbol{V}[/latex] and is given by [latex]\boldsymbol{Five = \textbf{emf} - Ir}[/latex], where [latex]\boldsymbol{I}[/latex] is the current and is positive when flowing abroad from the positive last of the voltage source.
• When multiple voltage sources are in series, their internal resistances add together and their emfs add algebraically.
• Solar cells can be wired in series or parallel to provide increased voltage or current, respectively.

Conceptual Questions

i: Is every emf a potential difference? Is every potential difference an emf? Explain.

2: Explain which bombardment is doing the charging and which is being charged in Figure 14.

Effigy xiv.

3: Given a bombardment, an assortment of resistors, and a variety of voltage and electric current measuring devices, describe how you would determine the internal resistance of the battery.

4: Two dissimilar 12-V automobile batteries on a store shelf are rated at 600 and 850 "cold cranking amps." Which has the smallest internal resistance?

5: What are the advantages and disadvantages of connecting batteries in serial? In parallel?

6: Semitractor trucks use four large 12-V batteries. The starter organisation requires 24 V, while normal operation of the truck'due south other electric components utilizes 12 V. How could the 4 batteries be connected to produce 24 V? To produce 12 V? Why is 24 5 better than 12 V for starting the truck's engine (a very heavy load)?

Trouble Exercises

one: Standard automobile batteries have six lead-acid cells in serial, creating a total emf of 12.0 5. What is the emf of an individual lead-acid jail cell?

2: Carbon-zinc dry cells (sometimes referred to as non-alkaline cells) accept an emf of 1.54 V, and they are produced as single cells or in diverse combinations to form other voltages. (a) How many one.54-5 cells are needed to make the mutual nine-Five battery used in many small electronic devices? (b) What is the actual emf of the approximately nine-V battery? (c) Discuss how internal resistance in the series connexion of cells volition affect the final voltage of this approximately 9-V battery.

three: What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell'due south internal resistance is [latex]\boldsymbol{2.00 \;\Omega}[/latex]?

4: (a) What is the terminal voltage of a big 1.54-V carbon-zinc dry prison cell used in a physics lab to supply two.00 A to a excursion, if the prison cell'southward internal resistance is [latex]\boldsymbol{0.100 \;\Omega}[/latex]? (b) How much electrical power does the cell produce? (c) What ability goes to its load?

5: What is the internal resistance of an automobile battery that has an emf of 12.0 V and a terminal voltage of 15.0 V while a electric current of 8.00 A is charging it?

half-dozen: (a) Observe the concluding voltage of a 12.0-V motorcycle battery having a [latex]\boldsymbol{0.600 - \;\Omega}[/latex] internal resistance, if it is being charged by a current of x.0 A. (b) What is the output voltage of the bombardment charger?

7: A car bombardment with a 12-V emf and an internal resistance of [latex]\boldsymbol{0.050 \;\Omega}[/latex] is being charged with a current of 60 A. Annotation that in this process the battery is being charged. (a) What is the potential divergence across its terminals? (b) At what charge per unit is thermal energy being dissipated in the battery? (c) At what charge per unit is electric energy being converted to chemical free energy? (d) What are the answers to (a) and (b) when the battery is used to supply 60 A to the starter motor?

eight: The hot resistance of a flashlight bulb is [latex]\boldsymbol{two.30 \;\Omega}[/latex], and it is run by a one.58-V alkaline jail cell having a [latex]\boldsymbol{0.100 - \;\Omega}[/latex] internal resistance. (a) What current flows? (b) Summate the power supplied to the bulb using [latex]\boldsymbol{I^2R_{\textbf{bulb}}}[/latex]. (c) Is this ability the same as calculated using [latex]\boldsymbol{\frac{V^2}{R_{\textbf{seedling}}}}[/latex]?

nine: The label on a portable radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-Five emf while alkaline cells have a one.58-V emf. The radio has a [latex]\boldsymbol{3.20 - \;\Omega}[/latex] resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio. (b) When using Nicad cells each having an internal resistance of [latex]\boldsymbol{0.0400 \;\Omega}[/latex]. (c) When using element of group i cells each having an internal resistance of [latex]\boldsymbol{0.200 \;\Omega}[/latex]. (d) Does this difference seem pregnant, considering that the radio's effective resistance is lowered when its volume is turned up?

10: An automobile starter motor has an equivalent resistance of [latex]\boldsymbol{0.0500 \;\Omega}[/latex] and is supplied by a 12.0-V battery with a [latex]\boldsymbol{0.0100 - \;\Omega}[/latex] internal resistance. (a) What is the electric current to the motor? (b) What voltage is applied to it? (c) What ability is supplied to the motor? (d) Repeat these calculations for when the bombardment connections are corroded and add [latex]\boldsymbol{0.0900 \;\Omega}[/latex] to the circuit. (Significant problems are caused by even pocket-sized amounts of unwanted resistance in depression-voltage, high-current applications.)

11: A kid's electronic toy is supplied by three 1.58-Five alkaline cells having internal resistances of [latex]\boldsymbol{0.0200 \;\Omega}[/latex] in series with a 1.53-V carbon-zinc dry cell having a [latex]\boldsymbol{0.100 - \;\Omega}[/latex] internal resistance. The load resistance is [latex]\boldsymbol{x.0 \;\Omega}[/latex]. (a) Draw a circuit diagram of the toy and its batteries. (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry out jail cell if it goes bad, resulting in merely 0.500 W existence supplied to the load?

12: (a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 5 when the current supplied increases by v.00 A? (b) Tin the emf of the voltage source be institute with the information supplied?

thirteen: A person with body resistance betwixt his hands of [latex]\boldsymbol{x.0 \;\textbf{k} \Omega}[/latex] accidentally grasps the terminals of a 20.0-kV ability supply. (Exercise NOT do this!) (a) Draw a circuit diagram to correspond the state of affairs. (b) If the internal resistance of the power supply is [latex]\boldsymbol{2000 \;\Omega}[/latex], what is the current through his body? (c) What is the ability dissipated in his body? (d) If the power supply is to exist made condom past increasing its internal resistance, what should the internal resistance be for the maximum current in this state of affairs to be i.00 mA or less? (e) Will this modification compromise the effectiveness of the power supply for driving low-resistance devices? Explain your reasoning.

14: Electric fish generate current with biological cells chosen electroplaques, which are physiological emf devices. The electroplaques in the South American eel are bundled in 140 rows, each row stretching horizontally forth the trunk and each containing 5000 electroplaques. Each electroplaque has an emf of 0.fifteen Five and internal resistance of [latex]\boldsymbol{0.25 \;\Omega}[/latex]. If the water surrounding the fish has resistance of [latex]\boldsymbol{800 \;\Omega}[/latex], how much current can the eel produce in water from near its head to most its tail?

15: Integrated Concepts

A 12.0-V emf automobile battery has a terminal voltage of sixteen.0 5 when being charged by a current of 10.0 A. (a) What is the battery's internal resistance? (b) What ability is dissipated inside the battery? (c) At what charge per unit (in ºC/min) volition its temperature increment if its mass is twenty.0 kg and information technology has a specific heat of [latex]\boldsymbol{0.300 \;\textbf{kcal/kg} \cdot ^{\circ} \textbf{C}}[/latex], assuming no oestrus escapes?

16: Unreasonable Results

A 1.58-V alkaline cell with a [latex]\boldsymbol{0.200 - \;\Omega}[/latex] internal resistance is supplying 8.50 A to a load. (a) What is its terminal voltage? (b) What is the value of the load resistance? (c) What is unreasonable most these results? (d) Which assumptions are unreasonable or inconsistent?

17: Unreasonable Results

(a) What is the internal resistance of a one.54-5 dry cell that supplies 1.00 W of power to a [latex]\boldsymbol{15.0 - \;\Omega}[/latex] bulb? (b) What is unreasonable well-nigh this result? (c) Which assumptions are unreasonable or inconsistent?

Glossary

electromotive forcefulness (emf)

the potential difference of a source of electricity when no current is flowing; measured in volts

internal resistance

the amount of resistance within the voltage source

potential deviation

the difference in electric potential between two points in an electrical circuit, measured in volts

concluding voltage

the voltage measured across the terminals of a source of potential difference

Solutions

Problems Exercises

1: 2.00 V

3: 2.9994 V

5: [latex]\boldsymbol{0.375 \;\Omega}[/latex]

viii: (a) 0.658 A

(b) 0.997 W

(c) 0.997 W; yes

x: (a) 200 A

(b) 10.0 5

(c) 2.00 kW

(d) [latex]\boldsymbol{0.g \;\Omega}[/latex]; lxxx.0 A, 4.0 V, 320 West

12: (a) [latex]\boldsymbol{0.400 \;\Omega}[/latex]

(b) No, at that place is only one independent equation, so only [latex]\boldsymbol{r}[/latex] can exist found.

16: (a) –0.120 V

(b) [latex]\boldsymbol{-ane.41 \times 10^{-two} \;\Omega}[/latex]

(c) Negative final voltage; negative load resistance.

(d) The assumption that such a cell could provide eight.l A is inconsistent with its internal resistance.

How To Find Terminal Voltage,

Source: http://pressbooks-dev.oer.hawaii.edu/collegephysics/chapter/21-2-electromotive-force-terminal-voltage/

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